1 Introduction
Throughout the paper p will denote a prime number, and $\mathbb {K}$ a field containing a root of unity of order p. Let $\mathbb {K}(p)$ denote the compositum of all finite Galois pextensions of $\mathbb {K}$ . The maximal prop Galois group of $\mathbb {K}$ , denoted by $G_{\mathbb {K}}(p)$ , is the Galois group $\operatorname {\mathrm {Gal}}(\mathbb {K}(p)/\mathbb {K})$ , and it coincides with the maximal prop quotient of the absolute Galois group of ${\mathbb {K}}$ . Characterising maximal prop Galois groups of fields among prop groups is one of the most important—and challenging—problems in Galois theory. One of the obstructions for the realization of a prop group as maximal prop Galois group for some field $\mathbb {K}$ is given by the Artin–Scherier theorem: the only finite group realizable as $G_{\mathbb {K}}(p)$ is the cyclic group of order 2 (cf. [Reference Becker1]).
The proof of the celebrated BlochKato conjecture, completed by Rost and Voevodsky with Weibel’s “patch” (cf. [Reference Haesemeyer and Weibel12, Reference Voevodsky27, Reference Weibel29]) provided new tools to study absolute Galois groups of field and their maximal prop quotients (see, e.g., [Reference Chebolu, Efrat and Mináč2, Reference Chebolu, Mináč and Quadrelli3, Reference Quadrelli17, Reference Quadrelli and Weigel21]). In particular, the nowcalled Norm Residue Theorem implies that the $\mathbb {Z}/p$ cohomology algebra of a maximal prop Galois group $G_{\mathbb {K}}(p)$
with $\mathbb {Z}/p$ a trivial $G_{\mathbb {K}}(p)$ module and endowed with the cupproduct, is a quadratic algebra: i.e., all its elements of positive degree are combinations of products of elements of degree 1, and its defining relations are homogeneous relations of degree 2 (see Section 2.3). For instance, from this property one may recover the ArtinSchreier obstruction (see, e.g., [Reference Quadrelli17, Section 2]).
More recently, a formal version of Hilbert 90 for prop groups was employed to find further results on the structure of maximal prop Galois groups (see [Reference Efrat and Quadrelli9, Reference Quadrelli19, Reference Quadrelli and Weigel21]). A pair $\mathcal {G}=(G,\theta )$ consisting of a prop group G endowed with a continuous representation $\theta \colon G\to \mathrm {GL}_1(\mathbb {Z}_p)$ is called a pro p pair. For a prop pair $\mathcal {G}=(G,\theta )$ let $\mathbb {Z}_p(1)$ denote the continuous left Gmodule isomorphic to $\mathbb {Z}_p$ as an abelian prop group, with Gaction induced by $\theta $ (namely, $g.v=\theta (g)\cdot v$ for every $v\in \mathbb {Z}_p(1)$ ). The pair $\mathcal {G}$ is called a Kummerian pro p pair if the canonical map
is surjective for every $n\geq 1$ . Moreover the pair $\mathcal {G}$ is said to be a 1smooth prop pair if every closed subgroup H, endowed with the restriction $\theta \vert _H$ , gives rise to a Kummerian prop pair (see Definition 2.1). By Kummer theory, the maximal prop Galois group $G_{\mathbb {K}}(p)$ of a field $\mathbb {K}$ , together with the prop cyclotomic character $\theta _{\mathbb {K}}\colon G_{\mathbb {K}}(p)\to \mathrm {GL}_1(\mathbb {Z}_p)$ (induced by the action of $G_{\mathbb {K}}(p)$ on the roots of unity of order a ppower lying in $\mathbb {K}(p)$ ) gives rise to a 1smooth prop pair $\mathcal {G}_{\mathbb {K}}$ (see Theorem 2.8).
In [Reference De Clercq and Florence5]—driven by the pursuit of an “explicit” proof of the Bloch–Kato conjecture as an alternative to the proof by Voevodsky—De Clerq and Florence introduced the 1smoothness property, and formulated the socalled “Smoothness Conjecture”: namely, that it is possible to deduce the surjectivity of the norm residue homomorphism (which is acknowledged to be the “hard part” of the Bloch–Kato conjecture) from the fact that $G_{\mathbb {K}}(p)$ together with the prop cyclotomic character is a 1smooth prop pair (see [Reference De Clercq and Florence5, Conjecture 14.25] and [Reference Mináč, Pop, Topaz and Wickelgren15, Section 3.1.6], and Question 2.10).
In view of the Smoothness Conjecture, it is natural to ask which properties of maximal prop Galois groups of fields arise also for 1smooth prop pairs. For example, the Artin–Scherier obstruction does: the only finite pgroup which may complete into a 1smooth prop pair is the cyclic group $C_2$ of order 2, together with the nontrivial representation $\theta \colon C_2\to \{\pm 1\}\subseteq \mathrm {GL}_1(\mathbb {Z}_2)$ (see Example 2.9).
A prop pair $\mathcal {G}=(G,\theta )$ comes endowed with a distinguished closed subgroup: the $\theta $ center $Z(\mathcal {G})$ of $\mathcal {G}$ , defined by
This subgroup is abelian, and normal in G. In [Reference Engler and Koenigsmann10], Engler and Koenigsmann showed that if the maximal prop Galois group $G_{\mathbb {K}}(p)$ of a field $\mathbb {K}$ is not cyclic then it has a unique maximal normal abelian closed subgroup (i.e., one containing all normal abelian closed subgroups of $G_{\mathbb {K}}(p)$ ), which coincides with the $\theta _{\mathbb {K}}$ center $Z(\mathcal {G}_{\mathbb {K}})$ , and the short exact sequence of prop groups
splits. We prove a grouptheoretic analogue of Engler–Koenigsmann’s result for 1smooth prop groups.
Theorem 1.1 Let G be a torsionfree prop group, $G\not \simeq \mathbb {Z}_p$ , endowed with a representation $\theta \colon G\to \mathrm {GL}_1(\mathbb {Z}_p)$ such that $\mathcal {G}=(G,\theta )$ is a 1smooth prop pair. Then $Z(\mathcal {G})$ is the unique maximal normal abelian closed subgroup of G, and the quotient $G/Z(\mathcal {G})$ is a torsionfree prop group.
In [Reference Ware28], Ware proved the following result on maximal prop Galois groups of fields: if $G_{\mathbb {K}}(p)$ is solvable, then it is locally uniformly powerful, i.e., $G_{\mathbb {K}}(p)\simeq A\rtimes \mathbb {Z}_p$ , where A is a free abelian prop group, and the rightside factor acts by scalar multiplication by a unit of $\mathbb {Z}_p$ (see Section 3.1). We prove that the same property holds also for 1smooth prop groups.
Theorem 1.2 Let G be a solvable torsionfree prop group, endowed with a representation $\theta \colon G\to \mathrm {GL}_1(\mathbb {Z}_p)$ such that $\mathcal {G}=(G,\theta )$ is a 1smooth prop pair. Then G is locally uniformly powerful.
This gives a complete description of solvable torsionfree prop groups which may be completed into a 1smooth prop pair. Moreover, Theorem 1.2 settles the Smoothness Conjecture positively for the class of solvable prop groups.
Corollary 1.3 If $\mathcal {G}=(G,\theta )$ is a 1smooth prop pair with G solvable, then G is a Bloch–Kato prop group, i.e., the $\mathbb {Z}/p$ cohomology algebra of every closed subgroup of G is quadratic.
Remark 1.4 After the submission of this paper, Snopce and Tanushevski showed in [Reference Snopce and Tanushevski24] that Theorems 1.2–1.1 hold for a wider class of prop groups. A prop group is said to be Frattiniinjective if distinct finitely generated closed subgroups have distinct Frattini subgroups (cf. [Reference Snopce and Tanushevski24, Definition 1.1]). By [Reference Snopce and Tanushevski24, Theorem 1.11 and Corollary 4.3], a prop group which may complete into a 1smooth prop pair is Frattiniinjective. By [Reference Snopce and Tanushevski24, Theorem 1.4] a Frattiniinjective prop group has a unique maximal normal abelian closed subgroup, and by [Reference Snopce and Tanushevski24, Theorem 1.3] a Frattiniinjective prop group is solvable if, and only if, it is locally uniformly powerful.
A solvable prop group does not contain a free nonabelian closed subgroup. For Bloch–Kato prop groups—and thus in particular for maximal prop Galois groups of fields containing a root of unity of order p—Ware proved the following Tits’ alternative: either such a prop group contains a free nonabelian closed subgroup; or it is locally uniformly powerful (see [Reference Ware28, Corollary 1] and [Reference Quadrelli17, Theorem B]). We conjecture that the same phenomenon occurs for 1smooth prop groups.
Conjecture 1.5 Let G be a torsionfree prop group which may be endowed with a representation $\theta \colon G\to \mathrm {GL}_1(\mathbb {Z}_p)$ such that $\mathcal {G}=(G,\theta )$ is a 1smooth prop pair. Then either G is locally uniformly powerful, or G contains a closed nonabelian free prop group.
2 Cyclotomic prop pairs
Henceforth, every subgroup of a prop group will be tacitly assumed to be closed, and the generators of a subgroup will be intended in the topological sense.
In particular, for a prop group G and a positive integer n, $G^{p^n}$ will denote the closed subgroup of G generated by the $p^n$ th powers of all elements of G. Moreover, for two elements $g,h\in G$ , we set
and for two subgroups $H_1,H_2$ of G, $[H_1,H_2]$ will denote the closed subgroup of G generated by all commutators $[h,g]$ with $h\in H_1$ and $g\in H_2$ . In particular, $G'$ will denote the commutator subgroup $[G,G]$ of G, and the Frattini subgroup $G^p\cdot G'$ of G is denoted by $\Phi (G)$ . Finally, $d(G)$ will denote the minimal number of generatord of G, i.e., $d(G)=\dim (G/\Phi (G))$ as a $\mathbb {Z}/p$ vector space.
2.1 Kummerian prop pairs
Let $1+p\mathbb {Z}_p=\{1+p\lambda \mid \lambda \in \mathbb {Z}_p\}\subseteq \mathrm {GL}_1(\mathbb {Z}_p)$ denote the prop Sylow subgroup of the group of units of the ring of padic integers $\mathbb {Z}_p$ . A pair $\mathcal {G}=(G,\theta )$ consisting of a prop group G and a continuous homomorphism
is called a cyclotomic pro p pair, and the morphism $\theta $ is called an orientation of G (cf. [Reference Efrat7, Section 3] and [Reference Quadrelli and Weigel21]).
A cyclotomic prop pair $\mathcal {G}=(G,\theta )$ is said to be torsionfree if $\operatorname {\mathrm {Im}}(\theta )$ is torsionfree: this is the case if p is odd; or if $p=2$ and $\operatorname {\mathrm {Im}}(\theta )\subseteq 1+4\mathbb {Z}_2$ . Observe that a cyclotomic prop pair $\mathcal {G}=(G,\theta )$ may be torsionfree even if G has nontrivial torsion—e.g., if G is the cyclic group of order p and $\theta $ is constantly equal to 1. Given a cyclotomic prop pair $\mathcal {G}=(G,\theta )$ one has the following constructions:

(a) if H is a subgroup of G, $\operatorname {\mathrm {Res}}_H(\mathcal {G})=(H,\theta \vert _H)$ ;

(b) if N is a normal subgroup of G contained in $\operatorname {\mathrm {Ker}}(\theta )$ , then $\theta $ induces an orientation $\bar \theta \colon G/N\to 1+p\mathbb {Z}_p$ , and we set $\mathcal {G}/N=(G/N,\bar \theta )$ ;

(c) if A is an abelian prop group, we set $A\rtimes \mathcal {G}=(A\rtimes G,\theta \circ \pi )$ , with $a^g=a^{\theta (g)^{1}}$ for all $a\in A$ , $g\in G$ , and $\pi $ the canonical projection $A\rtimes G\to G$ .
Given a cyclotomic prop pair $\mathcal {G}=(G,\theta )$ , the prop group G has two distinguished subgroups:

(a) the subgroup
(2.1) $$ \begin{align} K(\mathcal{G})=\left\langle\left. h^{\theta(g)}\cdot h^{g^{1}}\rightg\in G,h\in\operatorname{\mathrm{Ker}}(\theta)\right\rangle \end{align} $$introduced in [Reference Efrat and Quadrelli9, Section 3]; 
(b) the $\theta $ center
(2.2) $$ \begin{align} Z(\mathcal{G})=\left\langle h\in\operatorname{\mathrm{Ker}}(\theta)\leftghg^{1}=h^{\theta(g)}\;\forall\:g\in G\right.\right\rangle \end{align} $$introduced in [Reference Quadrelli17, Section 1].
Both $Z(\mathcal {G})$ and $K(\mathcal {G})$ are normal subgroups of G, and they are contained in $\operatorname {\mathrm {Ker}}(\theta )$ . Moreover, $Z(\mathcal {G})$ is abelian, while
Thus, the quotient $\operatorname {\mathrm {Ker}}(\theta )/K(\mathcal {G})$ is abelian, and if $\mathcal {G}$ is torsionfree one has an isomorphism of prop pairs
namely, $G/K(\mathcal {G})\simeq (\operatorname {\mathrm {Ker}}(\theta )/K(\mathcal {G}))\rtimes (G/\operatorname {\mathrm {Ker}}(\theta ))$ (where the action is induced by $\theta $ , in the latter), and both prop groups are endowed with the orientation induced by $\theta $ (cf. [Reference Quadrelli18, Equation 2.6]).
Definition 2.1 Given a cyclotomic prop pair $\mathcal {G}=(G,\theta )$ , let $\mathbb {Z}_p(1)$ denote the continuous Gmodule of rank 1 induced by $\theta $ , i.e., $\mathbb {Z}_p(1)\simeq \mathbb {Z}_p$ as abelian prop groups, and $g.\lambda =\theta (g)\cdot \lambda $ for every $\lambda \in \mathbb {Z}_p(1)$ . The pair $\mathcal {G}$ is said to be Kummerian if for every $n\geq 1$ the map
induced by the epimorphism of Gmodules $\mathbb {Z}_p(1)/p^n\to \mathbb {Z}_p(1)/p$ , is surjective. Moreover, $\mathcal {G}$ is 1smooth if $\operatorname {\mathrm {Res}}_H(\mathcal {G})$ is Kummerian for every subgroup $H\subseteq G$ .
Observe that the action of G on $\mathbb {Z}_p(1)/p$ is trivial, as $\operatorname {\mathrm {Im}}(\theta )\subseteq 1+p\mathbb {Z}_p$ . We say that a prop group G may complete into a Kummerian, or 1smooth, prop pair if there exists an orientation $\theta \colon G\to 1+p\mathbb {Z}_p$ such that the pair $(G,\theta )$ is Kummerian, or 1smooth.
Kummerian prop pairs and 1smooth prop pairs were introduced in [Reference Efrat and Quadrelli9] and in [Reference De Clercq and Florence5, Section 14] respectively. In [Reference Quadrelli and Weigel21], if $\mathcal {G}=(G,\theta )$ is a 1smooth prop pair, the orientation $\theta $ is said to be 1cyclotomic. Note that in [Reference De Clercq and Florence5, Section 14.1], a prop pair is defined to be 1smooth if the maps (2.4) are surjective for every open subgroup of G, yet by a limit argument this implies also that the maps (2.4) are surjective also for every closed subgroup of G (cf. [Reference Quadrelli and Weigel21, Corollary 3.2]).
Remark 2.1 Let $\mathcal {G}=(G,\theta )$ be a cyclotomic prop pair. Then $\mathcal {G}$ is Kummerian if, and only if, the map
induced by the epimorphism of continuous left Gmodules $\mathbb {Z}_p(1)\twoheadrightarrow \mathbb {Z}_p(1)/p$ , is surjective (cf. [Reference Quadrelli and Weigel21, Proposition 2.1])—here $H_{\mathrm {cts}}^*$ denotes continuous cochain cohomology as introduced by Tate in [Reference Tate26].
One has the following grouptheoretic characterization of Kummerian torsionfree prop pairs (cf. [Reference Efrat and Quadrelli9, Theorems 5.6 and 7.1] and [Reference Quadrelli20, Theorem 1.2]).
Proposition 2.2 A torsionfree cyclotomic prop pair $\mathcal {G}=(G,\theta )$ is Kummerian if and only if $\operatorname {\mathrm {Ker}}(\theta )/K(\mathcal {G})$ is a free abelian prop group.
Remark 2.3 Let $\mathcal {G}=(G,\theta )$ be a cyclotomic prop pair with $\theta \equiv \mathbf {1}$ , i.e., $\theta $ is constantly equal to 1. Since $K(\mathcal {G})=G'$ in this case, $\mathcal {G}$ is Kummerian if and only if the quotient $G/G'$ is torsionfree. Hence, by Proposition 2.2, $\mathcal {G}$ is 1smooth if and only if $H/H'$ is torsionfree for every subgroup $H\subseteq G$ . Prop groups with such property are called absolutely torsionfree, and they were introduced by Würfel in [Reference Würfel30]. In particular, if $\mathcal {G}=(G,\theta )$ is a 1smooth prop pair (with $\theta $ nontrivial), then $\operatorname {\mathrm {Res}}_{\operatorname {\mathrm {Ker}}(\theta )}(\mathcal {G})=(\operatorname {\mathrm {Ker}}(\theta ),\mathbf {1})$ is again 1smooth, and thus $\operatorname {\mathrm {Ker}}(\theta )$ is absolutely torsionfree. Hence, a prop group which may complete into a 1smooth prop pair is an absolutely torsionfreebycyclic prop group.
Example 2.4

(a) A cyclotomic prop pair $(G,\theta )$ with G a free prop group is 1smooth for any orientation $\theta \colon G\to 1+p\mathbb {Z}_p$ (cf. [Reference Quadrelli and Weigel21, Section 2.2]).

(b) A cyclotomic prop pair $(G,\theta )$ with G an infinite Demushkin prop group is 1smooth if and only if $\theta \colon G\to 1+p\mathbb {Z}_p$ is defined as in [Reference Labute14, Theorem 4] (cf. [Reference Efrat and Quadrelli9, Theorem 7.6]). E.g., if G has a minimal presentation
$$ \begin{align*} G=\left\langle\:x_1,\ldots,x_d\:\mid\:x_1^{p^f}[x_1,x_2]\cdots[x_{d1},x_d]=1\:\right\rangle \end{align*} $$with $f\geq 1$ (and $f\geq 2$ if $p=2$ ), then $\theta (x_2)=(1p^f)^{1}$ , while $\theta (x_i)=1$ for $i\neq 2$ . 
(c) For $p\neq 2$ let G be the prop group with minimal presentation
$$ \begin{align*}G=\langle x,y,z\mid [x,y]=z^p\rangle.\end{align*} $$Then the prop pair $(G,\theta )$ is not Kummerian for any orientation $\theta \colon G\to 1+p\mathbb {Z}_p$ (cf. [Reference Efrat and Quadrelli9, Theorem 8.1]). 
(d) Let
$$ \begin{align*}H=\left\{\left(\begin{array}{ccc} 1 & a & c \\ 0 & 1 & b \\ 0 & 0 &1 \end{array}\right)\mid a,b,c\in\mathbb{Z}_p\right\}\end{align*} $$be the Heisenberg prop group. The pair $(H,\mathbf {1})$ is Kummerian, as $H/H'\simeq \mathbb {Z}_p^2$ , but H is not absolutely torsionfree. In particular, H can not complete into a 1smooth prop pair (cf. [Reference Quadrelli18, Example 5.4]). 
(e) The only 1smooth prop pair $(G,\theta )$ with G a finite pgroup is the cyclic group of order 2 $G\simeq \mathbb {Z}/2$ , endowed with the only nontrivial orientation $\theta \colon G\twoheadrightarrow \{\pm 1\}\subseteq 1+2\mathbb {Z}_2$ (cf. [Reference Efrat and Quadrelli9, Example 3.5]).
Remark 2.5 By Example 2.4(e), if $\mathcal {G}=(G,\theta )$ is a torsionfree 1smooth prop pair, then G is torsionfree.
A torsionfree prop pair $\mathcal {G}=(G,\theta )$ is said to be $\theta $ abelian if the following equivalent conditions hold:

(i) $\operatorname {\mathrm {Ker}}(\theta )$ is a free abelian prop group, and $\mathcal {G}\simeq \operatorname {\mathrm {Ker}}(\theta )\rtimes (\mathcal {G}/\operatorname {\mathrm {Ker}}(\theta ))$ ;

(ii) $Z(\mathcal {G})$ is a free abelian prop group, and $Z(\mathcal {G})=\operatorname {\mathrm {Ker}}(\theta )$ ;

(iii) $\mathcal {G}$ is Kummerian and $K(\mathcal {G})=\{1\}$
(cf. [Reference Quadrelli17, Proposition 3.4] and [Reference Quadrelli20, Section 2.3]). Explicitly, a torsionfree prop pair $\mathcal {G}=(G,\theta )$ is $\theta $ abelian if and only if G has a minimal presentation
for some set I and some ppower q (possibly $q=p^\infty =0$ ), and in this case $\operatorname {\mathrm {Im}}(\theta )=1+q\mathbb {Z}_p$ . In particular, a $\theta $ abelian prop pair is also 1smooth, as every open subgroup U of G is again isomorphic to $\mathbb {Z}_p^I\rtimes \mathbb {Z}_p$ , with action induced by $\theta \vert _U$ , and therefore $\operatorname {\mathrm {Res}}_U(\mathcal {G})$ is $\theta \vert _U$ abelian.
Remark 2.6 From [Reference Efrat and Quadrelli9, Theorem 5.6], one may deduce also the following grouptheoretic characterization of Kummerian prop pairs: a prop group G may complete into a Kummerian oriented prop group if, and only if, there exists an epimorphism of prop groups $\varphi \colon G\twoheadrightarrow \bar G$ such that $\bar G$ has a minimal presentation (2.5), and $\operatorname {\mathrm {Ker}}(\varphi )$ is contained in the Frattini subgroup of G (cf., e.g., [Reference Quadrelli and Weigel22, Proposition 3.11]).
Remark 2.7 If $G\simeq \mathbb {Z}_p$ , then the pair $(G,\theta )$ is $\theta $ abelian, and thus also 1smooth, for any orientation $\theta \colon G\to 1+p\mathbb {Z}_p$ .
On the other hand, if $\mathcal {G}=(G,\theta )$ is a $\theta $ abelian prop pair with $d(G)\geq 2$ , then $\theta $ is the only orientation which may complete G into a 1smooth prop pair. Indeed, let $\mathcal {G}'=(G,\theta ')$ be a cyclotomic prop pair, with $\theta '\colon G\to 1+p\mathbb {Z}_p$ different to $\theta $ , and let $\{x_0,x_i,i\in I\}$ be a minimal generating set of G as in the presentation (2.5)—thus, $\theta (x_i)=1$ for all $i\in I$ , and $\theta (x_0)\in 1+q\mathbb {Z}_p$ . Then for some $i\in I$ one has $\theta '\vert _H\not \equiv \theta \vert _H$ , with H the subgroup of G generated by the two elements $x_0$ and $x_i$ . In particular, one has $\theta ([x_0,x_i])=\theta '([x_0,x_i])=1$ .
Suppose that $\mathcal {G}'$ is 1smooth. If $\theta '(x_i)\neq 1$ , then
hence $x_i^{q(1\theta '(x_i))}=1$ , a contradiction as G is torsionfree by Remark 2.5. If $\theta '(x_i)=1$ then necessarily $\theta '(x_0)\neq \theta (x_0)$ , and thus
hence $x_i^{\theta (x_0)\theta '(x_0)}=1$ , again a contradiction as G is torsionfree. (See also [Reference Quadrelli and Weigel21, Corollary 3.4].)
2.2 The Galois case
Let $\mathbb {K}$ be a field containing a root of 1 of order p, and let $\mu _{p^\infty }$ denote the group of roots of 1 of order a ppower contained in the separable closure of $\mathbb {K}$ . Then $\mu _{p^\infty }\subseteq \mathbb {K}(p)$ , and the action of the maximal prop Galois group $G_{\mathbb {K}}(p)=\operatorname {\mathrm {Gal}}(\mathbb {K}(p)/\mathbb {K})$ on $\mu _{p^\infty }$ induces a continuous homomorphism
—called the pro p cyclotomic character of $G_{\mathbb {K}}(p)$ —as the group of the automorphisms of $\mu _{p^{\infty }}$ which fix the roots of order p is isomorphic to $1+p\mathbb {Z}_p$ (see, e.g., [Reference Efrat8, p. 202] and [Reference Efrat and Quadrelli9, Section 4]). In particular, if $\mathbb {K}$ contains a root of 1 of order $p^k$ for $k\geq 1$ , then $\operatorname {\mathrm {Im}}(\theta _{\mathbb {K}})\subseteq 1+p^k\mathbb {Z}_p$ .
Set $\mathcal {G}_{\mathbb {K}}=(G_{\mathbb {K}}(p),\theta _{\mathbb {K}})$ . Then by Kummer theory one has the following (see, e.g., [Reference Efrat and Quadrelli9, Theorem 4.2]).
Theorem 2.8 Let $\mathbb {K}$ be a field containing a root of 1 of order p. Then $\mathcal {G}_{\mathbb {K}}=(G_{\mathbb {K}}(p),\theta _{\mathbb {K}})$ is 1smooth.
1smooth prop pairs share the following properties with maximal prop Galois groups of fields.
Example 2.9

(a) The only finite pgroup which occurs as maximal prop Galois group for some field $\mathbb {K}$ is the cyclic group of order 2, and this follows from the prop version of the Artin–Schreier Theorem (cf. [Reference Becker1]). Likewise, the only finite pgroup which may complete into a 1smooth prop pair, is the cyclic group of order 2 (endowed with the only nontrivial orientation onto $\{\pm 1\}$ ), as it follows from Example 2.4(e) and Remark 2.5.

(b) If x is an element of $G_{\mathbb {K}}(2)$ for some field $\mathbb {K}$ and x has order 2, then x selfcentralizes (cf. [Reference Craven and Smith4, Proposition 2.3]). Likewise, if x is an element of a pro $2$ group G which may complete into a 1smooth pro2 pair, then x selfcentralizes (cf. [Reference Quadrelli and Weigel21, Section 6.1]).
2.3 Bloch–Kato and the Smoothness Conjecture
A nonnegatively graded algebra $A_\bullet =\bigoplus _{n\geq 0}A_n$ over a field $\mathbb {F}$ , with $A_0=\mathbb {F}$ , is called a quadratic algebra if it is onegenerated—i.e., every element is a combination of products of elements of degree 1—and its relations are generated by homogeneous relations of degree 2. One has the following definitions (cf. [Reference De Clercq and Florence5, Definition 14.21] and [Reference Quadrelli17, Section 1]).
Definition 2.2 Let G be a prop group, and let $n\geq 1$ . Cohomology classes in the image of the natural cupproduct
are called symbols (relative to $\mathbb {Z}/p$ , wieved as trivial Gmodule).

(i) If for every open subgroup $U\subseteq G$ every element $\alpha \in H^n(U,\mathbb {Z}/p)$ , for every $n\geq 1$ , can be written as
$$ \begin{align*}\alpha=\mathrm{cor}_{V_1,U}^n(\alpha_1)+\cdots+\mathrm{cor}^n_{V_r,U}(\alpha_r),\end{align*} $$with $r\geq 1$ , where $\alpha _i\in H^n(V_i,\mathbb {Z}/p)$ is a symbol and$$ \begin{align*}\mathrm{cor}_{V_i,U}^n\colon H^n(V_i,\mathbb{Z}/p)\longrightarrow H^n(U,\mathbb{Z}/p)\end{align*} $$is the corestriction map (cf. [Reference Neukirch, Schmidt and Wingberg16, Chapter I, Section 5]), for some open subgroups $V_i\subseteq U$ , then G is called a weakly Bloch–Kato pro p group. 
(ii) If for every closed subgroup $H\subseteq G$ the $\mathbb {Z}/p$ cohomology algebra
$$ \begin{align*}H^\bullet(H,\mathbb{Z}/p)=\bigoplus_{n\geq0}H^n(H,\mathbb{Z}/p),\end{align*} $$endowed with the cupproduct, is a quadratic algebra over $\mathbb {Z}/p$ , then G is called a Bloch–Kato pro p group. As the name suggests, a Bloch–Kato prop group is also weakly BlochKato.
By the Norm Residue Theorem, if $\mathbb {K}$ contains a root of unity of order p, then the maximal prop Galois group $G_{\mathbb {K}}(p)$ is Bloch–Kato. The prop version of the “Smoothness Conjecture,” formulated by De Clerq and Florence, states that being 1smooth is a sufficient condition for a prop group to be weakly Bloch–Kato (cf. [Reference De Clercq and Florence5, Conjugation 14.25]).
Conjecture 2.10 Let $\mathcal {G}=(G,\theta )$ be a 1smooth prop pair. Then G is weakly Bloch–Kato.
In the case of $\mathcal {G}=\mathcal {G}_{\mathbb {K}}$ for some field $\mathbb {K}$ containing a root of 1 of order p, using Milnor Ktheory one may show that the weak Bloch–Kato condition implies that $H^\bullet (G,\mathbb {Z}/p)$ is onegenerated (cf. [Reference De Clercq and Florence5, Rem. 14.26]). In view of Theorem 2.8, a positive answer to the Smoothness Conjecture would provide a new proof of the surjectivity of the norm residue isomorphism, i.e., the “surjectivity” half of the Bloch–Kato conjecture (cf. [Reference De Clercq and Florence5, Section 1.1]).
Conjecture 2.10 has been settled positively for the following classes of prop groups.

(a) Finite pgroups: indeed, if $\mathcal {G}=(G,\theta )$ is a 1smooth prop pair with G a finite (nontrivial) pgroup, then by Example 2.4–(e) $p=2$ , G is a cyclic group of order two and $\theta \colon G\twoheadrightarrow \{\pm 1\}$ , so that $\mathcal {G}\simeq (\operatorname {\mathrm {Gal}}(\mathbb {C}/\mathbb {R}),\theta _{\mathbb {R}})$ , and G is Bloch–Kato.

(b) Analytic prop groups: indeed if $\mathcal {G}=(G,\theta )$ is a 1smooth prop pair with G a padic analytic prop group, then by [Reference Quadrelli18, Theorem 1.1] G is locally uniformly powerful and thus Bloch–Kato (see § 3.1 below).

(c) Prop completions of rightangled Artin groups: indeed, in [Reference Snopce and Zalesskii25], it is shown that if $\mathcal {G}=(G,\theta )$ is a 1smooth prop pair with G the prop completion of a rightangled Artin group induced by a simplicial graph $\Gamma $ , then necessarily $\theta $ is trivial and $\Gamma $ has the diagonal property—namely, G may be constructed starting from free prop groups by iterating the following two operations: free prop products, and direct products with $\mathbb {Z}_p$ —and thus G is Bloch–Kato (cf. [Reference Snopce and Zalesskii25, Theorem 1.2]).
3 Normal abelian subgroups
3.1 Powerful prop groups
Definition 3.1 A finitely generated prop group G is said to be powerful if one has $G'\subseteq G^p$ , and also $G'\subseteq G^4$ if $p=2$ . A powerful prop group which is also torsionfree and finitely generated is called a uniformly powerful prop group.
For the properties of powerful and uniformly powerful prop groups, we refer to [Reference Dixon, du Sautoy, Mann and Segal6, Chapter 4].
A prop group whose finitely generated subgroups are uniformly powerful, is said to be locally uniformly powerful. As mentioned in Section 1, a prop group G is locally uniformly powerful if, and only if, G has a minimal presentation (2.5)—i.e., G is locally powerful if, and only if, there exists an orientation $\theta \colon G\to 1+p\mathbb {Z}_p$ such that $(G,\theta )$ is a torsionfree $\theta $ abelian prop pair (cf. [Reference Quadrelli17, Theorem A] and [Reference Chebolu, Mináč and Quadrelli3, Proposition 3.5]).
Therefore, a locally uniformly powerful prop group G comes endowed automatically with an orientation $\theta \colon G\to 1+p\mathbb {Z}_p$ such that $\mathcal {G}=(G,\theta )$ is a 1smooth prop pair. In fact, finitely generated locally uniformly powerful prop groups are precisely those uniformly powerful prop groups which may complete into a 1smooth prop pair (cf. [Reference Quadrelli18, Proposition 4.3]).
Proposition 3.1 Let $\mathcal {G}=(G,\theta )$ be a 1smooth torsionfree prop pair. If G is locally powerful, then $\mathcal {G}$ is $\theta $ abelian, and thus G is locally uniformly powerful.
It is wellknown that the $\mathbb {Z}/p$ cohomology algebra of a prop group G with minimal presentation (2.5) is the exterior $\mathbb {Z}/p$ algebra
—if $p=2$ then $\bigwedge _{n\geq 0} V$ is defined to be the quotient of the tensor algebra over $\mathbb {Z}/p$ generated by V by the twosided ideal generated by the elements $v\otimes v$ , $v\in V$ —so that $H^\bullet (G,\mathbb {Z}/p)$ is quadratic. Moreover, every subgroup $H\subseteq G$ is again locally uniformly powerful, and thus also $H^\bullet (H,\mathbb {Z}/p)$ is quadratic. Hence, a locally uniformly powerful prop group is Bloch–Kato.
3.2 Normal abelian subgroups of maximal prop Galois groups
Let $\mathbb {K}$ be a field containing a root of 1 of order p (and also $\sqrt {1}$ if $p=2$ ). In Galois theory, one has the following result, due to Engler et al. (cf. [Reference Engler and Nogueira11] and [Reference Engler and Koenigsmann10]).
Theorem 3.2 Let $\mathbb {K}$ be a field containing a root of 1 of order p (and also $\sqrt {1}$ if $p=2$ ), and suppose that the maximal prop Galois group $G_{\mathbb {K}}(p)$ of $\mathbb {K}$ is not isomorphic to $\mathbb {Z}_p$ . Then $G_{\mathbb {K}}(p)$ contains a unique maximal abelian normal subgroup.
By [Reference Quadrelli and Weigel21, Theorem 7.7], such a maximal abelian normal subgroup coincides with the $\theta _{\mathbb {K}}$ center $Z(\mathcal {G}_{\mathbb {K}})$ of the prop pair $\mathcal {G}_{\mathbb {K}}=(G_{\mathbb {K}}(p),\theta _{\mathbb {K}})$ induced by the prop cyclotomic character $\theta _{\mathbb {K}}$ (cf. § 2.2). Moreover, the field $\mathbb {K}$ admits a pHenselian valuation with residue characteristic not p and nonpdivisible value group, such that the residue field $\kappa $ of such a valuation gives rise to the cyclotomic prop pair $\mathcal {G}_{\kappa }$ isomorphic to $\mathcal {G}_{\mathbb {K}}/Z(\mathcal {G}_{\mathbb {K}})$ , and the induced short exact sequence of prop groups
splits (cf. [Reference Engler and Koenigsmann10, Section 1] and [Reference Efrat8, Example 22.1.6]—for the definitions related to phenselian valuations of fields, we direct the reader to [Reference Efrat8, Section 15.3]). In particular, $G_{\mathbb {K}}(p)/Z(\mathcal {G}_{\mathbb {K}})$ is torsionfree.
Remark 3.3 By [Reference Quadrelli and Weigel21, Theorems 1.2 and 7.7], Theorem 3.2 and the splitting of (3.1) generalize to 1smooth prop pairs whose underlying prop group is Bloch–Kato. Namely, if $\mathcal {G}=(G,\theta )$ is a 1smooth prop pair with G a Bloch–Kato prop group, then $Z(\mathcal {G})$ is the unique maximal abelian normal subgroup of G, and it has a complement in G.
3.3 Proof of Theorem 1.1
In order to prove Theorem 1.1 (and also Theorem 1.2 later on), we need the following result.
Proposition 3.4 Let $\mathcal {G}=(G,\theta )$ be a torsionfree 1smooth prop pair, with $d(G)=2$ and $G=\langle x,y\rangle $ . If $[[x,y],y]=1$ , then $\operatorname {\mathrm {Ker}}(\theta )=\langle y\rangle $ and
Proof Let H be the subgroup of G generated by y and $[x,y]$ . Recall that by Remark 2.5, G (and hence also H) is torsionfree.
If $d(H)=1$ then $H\simeq \mathbb {Z}_p$ , as H is torsionfree. Moreover, H is generated by y and $x^{1}yx$ , and thus $xHx^{1}\subseteq H$ . Therefore, x acts on $H\simeq \mathbb {Z}_p$ by multiplication by $1+p\lambda $ for some $\lambda \in \mathbb {Z}_p$ . If $\lambda =0$ then G is abelian, and thus $G\simeq \mathbb {Z}_p^2$ as it is absolutely torsionfree, and $\theta \equiv \mathbf {1}$ by Remark 2.7. If $\lambda \neq 0$ then x acts nontrivially on the elements of H, and thus $\langle x\rangle \cap H=\{1\}$ and $G=H\rtimes \langle x\rangle $ : by (2.5), $(G,\theta ')$ is a $\theta '$ abelian prop pair, with $\theta '\colon G\to 1+p\mathbb {Z}_p$ defined by $\theta '(x)=1+p\lambda $ and $\theta '(y)=1$ . By Remark 2.7, one has $\theta '\equiv \theta $ , and thus $\theta (x)=1+p\lambda $ and $\theta (y)=1$ .
If $d(H)=2$ , then H is abelian by hypothesis, and torsionfree, and thus $(H,\theta ')$ is $\theta '$ abelian, with $\theta '\equiv {\mathbf {1}}\colon H\to 1+p\mathbb {Z}_p$ trivial. By Remark 2.7, one has $\theta '=\theta \vert _H$ , and thus $y,[x,y]\in \operatorname {\mathrm {Ker}}(\theta )$ . Now put $z=[x,y]$ and $t=y^p$ , and let U be the open subgroup of G generated by $x,z,t$ . Clearly, $\operatorname {\mathrm {Res}}_U(\mathcal {G})$ is again 1smooth. By hypothesis one has $z^y=z$ , and hence commutator calculus yields
Put $\lambda =1\theta (x)^{1}\in p\mathbb {Z}_p$ . Since $t\in \operatorname {\mathrm {Ker}}(\theta )$ , by (2.1) $[x,t]\cdot t^{\lambda }$ lies in $K(\operatorname {\mathrm {Res}}_U(\mathcal {G}))$ . Since t and z commute, from (3.2) one deduces
Moreover, $zt^{\lambda /p}\in \operatorname {\mathrm {Ker}}(\theta \vert _U)$ . Since $\operatorname {\mathrm {Res}}_U(\mathcal {G})$ is 1smooth, by Proposition 2.2, the quotient $\operatorname {\mathrm {Ker}}(\theta \vert _U)/K(\operatorname {\mathrm {Res}}_U(\mathcal {G}))$ is a free abelian prop group, and therefore (3.3) implies that also $zt^{\lambda /p}$ is an element of $K(\operatorname {\mathrm {Res}}_U(\mathcal {G}))$ .
Since $K(\operatorname {\mathrm {Res}}_U(\mathcal {G}))\subseteq \Phi (U)$ , one has $z\equiv t^{\lambda /p}\bmod \Phi (U)$ . Then by [Reference Dixon, du Sautoy, Mann and Segal6, Proposition 1.9] $d(U)=2$ and U is generated by x and t. Since $[x,t]\in U^p$ by (3.2), the prop group U is powerful. Therefore, $\operatorname {\mathrm {Res}}_U(\mathcal {G})$ is $\theta \vert _U$ abelian by Proposition 3.1. In particular, the subgroup $K(\operatorname {\mathrm {Res}}_U(\mathcal {G}))$ is trivial, and thus
and the claim follows.▪
Proposition 3.4 is a generalization of [Reference Quadrelli18, Proposition 5.6].
Theorem 3.5 Let $\mathcal {G}=(G,\theta )$ be a torsionfree 1smooth prop pair, with $d(G)\geq 2$ .

(i) The $\theta $ center $Z(\mathcal {G})$ is the unique maximal abelian normal subgroup of G.

(ii) The quotient $G/Z(\mathcal {G})$ is a torsionfree prop group.
Proof Recall that G is torsionfree by Remark 2.5. Since $Z(\mathcal {G})$ is an abelian normal subgroup of G by definition, in order to prove (i) we need to show that if A is an abelian normal subgroup of G, then $A\subseteq Z(\mathcal {G})$ .
First, we show that $A\subseteq \operatorname {\mathrm {Ker}}(\theta )$ . If $A\simeq \mathbb {Z}_p$ , let y be a generator of A. For every $x\in G$ one has $xyx^{1}\in A$ , and thus $xyx^{1}=y^{\lambda }$ , for some $\lambda \in 1+p\mathbb {Z}_p$ . Let H be the subgroup of G generated by x and y, for some $x\in G$ such that $d(H)=2$ . Then the pair $(H,\theta ')$ is $\theta '$ abelian for some orientation $\theta '\colon H\to 1+p\mathbb {Z}_p$ such that $y\in \operatorname {\mathrm {Ker}}(\theta ')$ , as H has a presentation as in (2.5). Since both $\operatorname {\mathrm {Res}}_H(\mathcal {G})$ and $(H,\theta ')$ are 1smooth prop pairs, by Remark 2.7, one has $\theta '=\theta \vert _H$ , and thus $A\subseteq \operatorname {\mathrm {Ker}}(\theta )$ .
If $A\not \simeq \mathbb {Z}_p$ , then A is a free abelian prop group with $d(A)\geq 2$ , as G is torsionfree. Therefore, by Remark 2.3 the prop pair $(A,\mathbf {1})$ is 1smooth. Since also $\operatorname {\mathrm {Res}}_A(\mathcal {G})$ is 1smooth, Remark 2.7 implies that $\theta \vert _A=\mathbf {1}$ , and hence $A\subseteq \operatorname {\mathrm {Ker}}(\theta )$ .
Now, for arbitrary elements $x\in G$ and $y\in A$ , put $z=[x,y]$ . Since A is normal in G, one has $z\in A$ , and since A is abelian, one has $[z,y]=1$ . Then Proposition 3.4 applied to the subgroup of G generated by $\{x,y\}$ yields $xyx^{1}=x^{\theta (x)}$ , and this completes the proof of statement (i).
In order to prove statement (ii), suppose that $y^p\in Z(\mathcal {G})$ for some $y\in G$ . Then $y^p\in \operatorname {\mathrm {Ker}}(\theta )$ , and since $\operatorname {\mathrm {Im}}(\theta )$ has no nontrivial torsion, also y lies in $\operatorname {\mathrm {Ker}}(\theta )$ . Since G is torsionfree by Remark 2.5, $y^p\neq 1$ . Let H be the subgroup of G generated by y and x, for some $x\in G$ such that $d(H)\geq 2$ . Since $xy^px^{1}=(y^p)^{\theta (x)}$ , commutator calculus yields
Put $z=[x,y]$ , and let S be the subgroup of H generated by $y,z$ . Clearly, $\operatorname {\mathrm {Res}}_S(\mathcal {G})$ is 1smooth, and since $y,z\in \operatorname {\mathrm {Ker}}(\theta )$ , one has $\theta \vert _S=\mathbf {1}$ , and thus $S/S'$ is a free abelian prop group by Remark 2.3. From (3.4) one deduces
Since $S/S'$ is torsionfree, (3.5) implies that $z\equiv y^{1\theta (x)^{1}}\bmod \Phi (S)$ , so that S is generated by y, and $S\simeq \mathbb {Z}_p$ , as G is torsionfree. Therefore, $S'=\{1\}$ , and (3.5) yields $[x,y]=y^{1\theta (x)^{1}}$ , and this completes the proof of statement (ii).▪
Remark 3.6 Let G be a prop group isomorphic to $\mathbb {Z}_p$ , and let $\theta \colon G\to 1+p\mathbb {Z}_p$ be a nontrivial orientation. Then by Example 2.4(a), $\mathcal {G}=(G,\theta )$ is 1smooth. Since G is abelian and $\theta (x)\neq 1$ for every $x\in G$ , $x\neq 1$ , $Z(\mathcal {G})=\{1\}$ , still every subgroup of G is normal and abelian.
In view of the splitting of (3.1) (and in view of Remark 3.3), it seems natural to ask the following question.
Question 3.7 Let $\mathcal {G}=(G,\theta )$ be a torsionfree 1smooth prop pair, with $d(G)\geq 2$ . Is the prop pair $\mathcal {G}/Z(\mathcal {G})=(G/Z(\mathcal {G}),\bar {\theta })$ 1smooth? Does the short exact sequence of prop groups
split?
If $\mathcal {G}=(G,\theta )$ is a torsionfree prop pair, then either $\operatorname {\mathrm {Ker}}(\theta )=G$ , or $\operatorname {\mathrm {Im}}(\theta )\simeq \mathbb {Z}_p$ , hence in the former case one has $G\simeq \operatorname {\mathrm {Ker}}(\theta )\rtimes (G/\operatorname {\mathrm {Ker}}(\theta ))$ , as the rightside factor is isomorphic to $\mathbb {Z}_p$ , and thus pprojective (cf. [Reference Neukirch, Schmidt and Wingberg16, Chapter III, Section 5]). Since $Z(\mathcal {G})\subseteq Z(\operatorname {\mathrm {Ker}}(\theta ))$ (and $Z(\mathcal {G})= Z(G)$ if $\operatorname {\mathrm {Ker}}(\theta )=G$ ), and since $\operatorname {\mathrm {Ker}}(\theta )$ is absolutely torsionfree if $\mathcal {G}$ is 1smooth, Question 3.7 is equivalent to the following question (of its own grouptheoretic interest): if G is an absolutely torsionfree prop group, does G split as direct product
One has the following partial answer (cf. [Reference Würfel30, Proposition 5]): if G is absolutely torsionfree, and $Z(G)$ is finitely generated, then $\Phi _n(G)=Z(\Phi _n(G))\times H$ , for some $n\geq 1 $ and some subgroup $H\subseteq \Phi _n(G)$ (here $\Phi _n(G)$ denotes the iterated Frattini series of G, i.e., $\Phi _1(G)=G$ and $\Phi _{n+1}(G)=\Phi (\Phi _n(G))$ for $n\geq 1$ ).
4 Solvable prop groups
4.1 Solvable prop groups and maximal prop Galois groups
Recall that a (prop) group G is said to be metaabelian if there is a short exact sequence
such that both N and $\bar G$ are abelian; or, equivalently, if the commutator subgroup $G'$ is abelian. Moreover, a prop group G is solvable if the derived series $(G^{(n)})_{n\geq 1}$ of G—i.e., $G^{(1)}=G$ and $G^{(n+1)}=[G^{(n)},G^{(n)}]$ —is finite, namely $G^{(N+1)}=\{1\}$ for some finite N.
Example 4.1 A nonabelian locally uniformly powerful prop group G is metaabelian: if $\theta \colon G\to 1+p\mathbb {Z}_p$ is the associated orientation, then $G'\subseteq \operatorname {\mathrm {Ker}}(\theta )^p$ , and thus $G'$ is abelian.
In Galois theory, one has the following result by Ware (cf. [Reference Ware28, Theorem 3], see also [Reference Koenigsmann13] and [Reference Quadrelli17, Theorem 4.6]).
Theorem 4.2 Let $\mathbb {K}$ be a field containing a root of 1 of order p (and also $\sqrt {1}$ if $p=2$ ). If the maximal prop Galois group $G_{\mathbb {K}}(p)$ is solvable, then $\mathcal {G}_{\mathbb {K}}$ is $\theta _{\mathbb {K}}$ abelian.
4.2 Proof of Theorem 1.2 and Corollary 1.3
In order to prove Theorem 1.2, we prove first the following intermediate results—a consequence of Würfel’s result [Reference Würfel30, Proposition 2] —, which may be seen as the “1smooth analogue” of [Reference Ware28, Theorem 2].
Proposition 4.3 Let $\mathcal {G}=(G,\theta )$ be a torsionfree 1smooth prop pair. If G is metaabelian, then $\mathcal {G}$ is $\theta $ abelian.
Proof Assume first that $\theta \equiv \mathbf {1}$ —i.e., G is absolutely torsionfree (cf. Remark 2.3). Then G is a free abelian prop group by [Reference Würfel30, Proposition 2].
Assume now that $\theta \not \equiv \mathbf {1}$ . Since $\mathcal {G}$ is 1smooth, also $\operatorname {\mathrm {Res}}_{\operatorname {\mathrm {Ker}}(\theta )}(\mathcal {G})$ and $\operatorname {\mathrm {Res}}_{\operatorname {\mathrm {Ker}}(\theta )'}(\mathcal {G})$ are 1smooth prop pairs, and thus $\operatorname {\mathrm {Ker}}(\theta )$ and $\operatorname {\mathrm {Ker}}(\theta )'$ are absolutely torsionfree. Moreover, $\operatorname {\mathrm {Ker}}(\theta )'\subseteq G'$ , and since the latter is abelian, also $\operatorname {\mathrm {Ker}}(\theta )'$ is abelian, i.e., $\operatorname {\mathrm {Ker}}(\theta )$ is metaabelian. Thus $\operatorname {\mathrm {Ker}}(\theta )$ is a free abelian prop group by [Reference Würfel30, Proposition 2]. Consequently, for arbitrary $y\in \operatorname {\mathrm {Ker}}(\theta )$ and $x\in G$ , the commutator $[x,y]$ lies in $\operatorname {\mathrm {Ker}}(\theta )$ and $[[x,y],y]=1$ . Therefore, Proposition 3.4 implies that $xyx^{1}=y^{\theta (y)}$ for every $x\in G$ and $y\in \operatorname {\mathrm {Ker}}(\theta )$ , namely, $\mathcal {G}$ is $\theta $ abelian.▪
Note that Proposition 4.3 generalizes [Reference Würfel30, Proposition 2] from absolutely torsionfree prop groups to 1smooth prop groups. From Proposition 4.3, we may deduce Theorem 1.2.
Proposition 4.4 Let $\mathcal {G}=(G,\theta )$ be a torsionfree 1smooth prop pair. If G is solvable, then G is locally uniformly powerful.
Proof Let N be the positive integer such that $G^{(N)}\neq \{1\}$ and $G^{(N+1)}=\{1\}$ . Then for every $1\leq n\leq N$ , the prop pair $\operatorname {\mathrm {Res}}_{G^{n}}(\mathcal {G})$ is 1smooth, and $G^{(n)}$ is solvable, and moreover $\theta \vert _{G^{(n)}}\equiv \mathbf {1}$ if $n\geq 2$ .
Suppose that $N\geq 3$ . Since $G^{(N1)}$ is metaabelian and $\theta \vert _{G^{(N1)}}\equiv \mathbf {1}$ , Proposition 4.3 implies that $G^{(N1)}$ is a free abelian prop group, and therefore $G^{(N)}=\{1\}$ , a contradiction. Thus, $N\leq 2$ , and G is metaabelian. Therefore, Proposition 4.3 implies that the prop pair $\mathcal {G}$ is $\theta $ abelian, and hence G is locally uniformly powerful (cf. § 3.1).▪
Proposition 4.4 may be seen as the 1smooth analogue of Ware’s Theorem 4.2. Corollary 1.3 follows from Proposition 4.4 and from the fact that a locally uniformly powerful prop group is Bloch–Kato (cf. § 3.1).
Corollary 4.5 Let $\mathcal {G}=(G,\theta )$ be a torsionfree 1smooth prop pair. If G is solvable, then G is Bloch–Kato.
This settles the Smoothness Conjecture for the class of solvable prop groups.
4.3 A Tits’ alternative for 1smooth prop groups
For maximal prop Galois groups of fields one has the following Tits’ alternative (cf. [Reference Ware28, Corollary 1]).
Theorem 4.6 Let $\mathbb {K}$ be a field containing a root of 1 of order p (and also $\sqrt {1}$ if $p=2$ ). Then either $\mathcal {G}_{\mathbb {K}}$ is $\theta _{\mathbb {K}}$ abelian, or $G_{\mathbb {K}}(p)$ contains a closed nonabelian free prop group.
Actually, the above Tits’ alternative holds also for the class of Bloch–Kato prop groups, with p odd: if a Bloch–Kato prop group G does not contain any free nonabelian subgroups, then it can complete into a $\theta $ abelian prop pair $\mathcal {G}=(G,\theta )$ (cf. [Reference Quadrelli17, Theorem B], this Tits’ alternative holds also for $p=2$ under the further assumption that the Bockstein morphism $\beta \colon H^1(G,\mathbb {Z}/2)\to H^2(G,\mathbb {Z}/2)$ is trivial, see [Reference Quadrelli17, Theorem 4.11]).
Clearly, a solvable prop group contains no free nonabelian subgroups.
A prop group is padic analytic if it is a padic analytic manifold and the map $(x,y)\mapsto x^{1} y$ is analytic, or, equivalently, if it contains an open uniformly powerful subgroup (cf. [Reference Dixon, du Sautoy, Mann and Segal6, Theorem 8.32])—e.g., the Heisenberg prop group is analytic. Similarly to solvable prop groups, a padic analytic prop group does not contain a free nonabelian subgroup (cf. [Reference Dixon, du Sautoy, Mann and Segal6, Corollary 8.34]).
Even if there are several padic analytic prop groups which are solvable (e.g., finitely generated locally uniformly powerful prop groups), none of these two classes of prop groups contains the other one: e.g.,

(a) the wreath product $\mathbb {Z}_p\wr \mathbb {Z}_p\simeq \mathbb {Z}_p^{\mathbb {Z}_p}\rtimes \mathbb {Z}_p$ is a metaabelian prop group, but it is not padic analytic (cf. [Reference Shalev23]) and

(b) if G is a propSylow subgroup of $\mathrm {SL}_2(\mathbb {Z}_p)$ , then G is a padic analytic prop group, but it is not solvable.
In addition, it is wellknown that also for the class of prop completions of rightangled Artin prop groups one has a Tits’ alternative: the prop completion of a rightangled Artin prop group contains a free nonabelian subgroup unless it is a free abelian prop group (i.e., unless the associated graph is complete)—and thus it is locally uniformly powerful.
In [Reference Quadrelli18], it is shown that analytic prop groups which may complete into a 1smooth prop pair are locally uniformly powerful. Therefore, after the results in [Reference Quadrelli18] and [Reference Snopce and Zalesskii25], and Theorem 1.2, it is natural to ask whether a Tits’ alternative, analogous to Theorem 4.6 (and its generalization to Bloch–Kato prop groups), holds also for all torsionfree 1smooth prop pairs.
Question 4.7 Let $\mathcal {G}=(G,\theta )$ be a torsionfree 1smooth prop pair, and suppose that $\mathcal {G}$ is not $\theta $ abelian. Does G contain a closed nonabelian free prop group?
In other words, we are asking whether there exists torsionfree 1smooth prop pairs $\mathcal {G}=(G,\theta )$ such that G is not analytic nor solvable, and yet it contains no free nonabelian subgroups. In view of Theorem 4.6 and of the Tits’ alternative for Bloch–Kato prop groups [Reference Quadrelli17, Theorem B], a positive answer to Question 4.7 would corroborate the Smoothness Conjecture.
Observe that—analogously to Question 3.7—Question 4.7 is equivalent to asking whether an absolutely torsionfree prop group which is not abelian contains a closed nonabelian free subgroup. Indeed, by Proposition 3.4 (in fact, just by [Reference Quadrelli18, Proposition 5.6]), if $\mathcal {G}=(G,\theta )$ is a torsionfree 1smooth prop pair and $\operatorname {\mathrm {Ker}}(\theta )$ is abelian, then $\mathcal {G}$ is $\theta $ abelian.
Acknowledgment
The author thanks I. Efrat, J. Minac, N.D. Tân, and Th. Weigel for working together on maximal prop Galois groups and their cohomology; and P. Guillot and I. Snopce for the interesting discussions on 1smooth prop groups. Also, the author wishes to thank the editors of CMBBMC, for their helpfulness, and the anonymous referee.